Linear Algebra - set of piecewise continuous functions is a vector space

[corey115]

Homework Statement

A function f:[a,b] \rightarrow ℝ is called piecewise continuous if there exists a finite number of points a = x₀ < x₁ < x₂ < ... < x_k-1 < x_k = b such that
(a) f is continuous on (x_i-1, x_i) for i = 0, 1, 2, ..., k
(b) the one sided limits exist as finite numbers

Let V be the set of all piecewise continuous functions on [a, b]. Prove that V is a vector space over ℝ, with addition and scalar multiplication defined as usual for functions.

Homework Equations

The axioms for fields and vector spaces.

The Attempt at a Solution

If function addition and scalar multiplication work in the same way as usual, don't I just have to make an argument that the limits still exist as finite numbers (since our scalars are finite)? And a similar argument that if I add two functions that those limits will just be the sum of the two limits of each function?

 

[jbunniii]

If function addition and scalar multiplication work in the same way as usual, don't I just have to make an argument that the limits still exist as finite numbers (since our scalars are finite)? And a similar argument that if I add two functions that those limits will just be the sum of the two limits of each function?

Suppose f, g\in V, and let h = f + g. If f and g are both continuous at a point x, what can you say about h at that point? Is it continuous at x?

 

[corey115]

Suppose f, g\in V, and let h = f + g. If f and g are both continuous at a point x, what can you say about h at that point? Is it continuous at x?

It will also be continuous because the sum of two continuous functions will also be continuous. But do I need extra arguments besides that because V is the set of piecewise continuous functions?

 

[jbunniii]

It will also be continuous because the sum of two continuous functions will also be continuous. But do I need extra arguments besides that because V is the set of piecewise continuous functions?

Yes, but we have learned something already. h will be continuous at any point where both f and g are continuous, so h can only be discontinuous at points where f or g are discontinuous, and there are only finitely many such points. Thus h can have only finitely many discontinuities. So h satisfies condition (a).

Now consider condition (b). If h is discontinuous at x, what are \lim_{y \rightarrow x^+} h(x) and \lim_{y \rightarrow x^-} h(x)? Can you write these in terms of the corresponding one-sided limits of f and g?

 

[corey115]

Yes, but we have learned something already. h will be continuous at any point where both f and g are continuous, so h can only be discontinuous at points where f or g are discontinuous, and there are only finitely many such points. Thus h can have only finitely many discontinuities. So h satisfies condition (a).

Now consider condition (b). If h is discontinuous at x, what are \lim_{y \rightarrow x^+} h(x) and \lim_{y \rightarrow x^-} h(x)? Can you write these in terms of the corresponding one-sided limits of f and g?

Would that not be just the sum of the two limits of f and g at x, or if one or both of those two functions is discontinuous at x, then I would just use the whatever sided-limit that I need?

 

[jbunniii]

Would that not be just the sum of the two limits of f and g at x, or if one or both of those two functions is discontinuous at x, then I would just use the whatever sided-limit that I need?

Right, you have \lim_{y \rightarrow x^+} h(y) = \lim_{y \rightarrow x^+} f(y) + \lim_{y \rightarrow x^+} g(y), and similarly with the left hand limit. Since these limits exist for f and g, they also exist for h. This shows that h satisfies (b).

 

[jbunniii]

So that shows that V is closed under addition. What else do you need to show in order to prove that V is a vector space?

 

[corey115]

So that shows that V is closed under addition. What else do you need to show in order to prove that V is a vector space?

Closure under scalar multiplication.
Commutative/Associative for addition/multiplication.
Additive identity/inverse.
Multiplicative identity.
Distributive property.
Associative property for scalar multiplication.

 

[jbunniii]

Commutative/Associative for addition/multiplication.
Distributive property.
Associative property for scalar multiplication.

OK, these three are true for any functions so they remain true for piecewise continuous functions.

Additive identity/inverse.
Multiplicative identity.

Which functions do these correspond to? Are they piecewise continuous?

Closure under scalar multiplication.

If f \in V and c is a scalar, is cf piecewise continuous? Should be pretty easy to show that it is.

 

[corey115]

OK, these three are true for any functions so they remain true for piecewise continuous functions.

Which functions do these correspond to? Are they piecewise continuous?

If f \in V and c is a scalar, is cf piecewise continuous? Should be pretty easy to show that it is.

For the 2nd item you listed, would the additive inverse of f be -f and the additive identity of f be the 0 function that is discontinuous where ever f is discontinuous?

As for the 3rd item, I'm not entirely sure how I would go about showing that. It makes sense intuitively as to why it would be true, but I'm not sure how to put that on paper.

 

[jbunniii]

For the 2nd item you listed, would the additive inverse of f be -f

Yes. Note that -f is discontinuous at exactly the same points as f, and the left and right limits exist because they exist for f. (Insert easy proofs as needed.)

and the additive identity of f be the 0 function that is discontinuous where ever f is discontinuous?

The additive identity of V would be just the standard 0 function, which is continuous everywhere. Note that all continuous functions are in V because the number of discontinuities (zero) is certainly finite, and the left and right limits exist due to continuity.

As for the 3rd item, I'm not entirely sure how I would go about showing that. It makes sense intuitively as to why it would be true, but I'm not sure how to put that on paper.

Claim: cf is continuous at any point x where f is continuous. (Proof?) Therefore cf has at most finitely many discontinuities. What about \lim_{y \rightarrow x^+} cf(y)? What does this equal?

 

[corey115]

Yes. Note that -f is discontinuous at exactly the same points as f, and the left and right limits exist because they exist for f. (Insert easy proofs as needed.)

The additive identity of V would be just the standard 0 function, which is continuous everywhere. Note that all continuous functions are in V because the number of discontinuities (zero) is certainly finite, and the left and right limits exist due to continuity.

Claim: cf is continuous at any point x where f is continuous. (Proof?) Therefore cf has at most finitely many discontinuities. What about \lim_{y \rightarrow x^+} cf(y)? What does this equal?

Since that is just a constant (scalar in our case), it would equal c times whatever the limit of just f(x) was.

 

[jbunniii]

Since that is just a constant (scalar in our case), it would equal c times whatever the limit of just f(x) was.

Right. Expressing it formally:
$$\lim_{y \rightarrow x^+} cf(y) = c \lim_{y \rightarrow x^+}f(y)$$
by the linearity property of limits, and the right hand side exists because f \in V.

 

[corey115]

Right. Expressing it formally:
$$\lim_{y \rightarrow x^+} cf(y) = c \lim_{y \rightarrow x^+}f(y)$$
by the linearity property of limits, and the right hand side exists because f \in V.

You have been more than helpful! I feel like I have a much greater understanding of this proof now!